
\prob{00B1}{均值不等式}

对于实数$a_1, a_2, \dots, a_n$，求证：
\[ \sum_{i = 1}^n a_i \ge n\sqrt[n]{\prod_{i = 1}^n a_i} \]
当且仅当$a_1 = a_2 = \dots = a_n$时等号成立。
\problabels{yellow/代数, green/证明题}

\subsection{归纳法}

\subsubsection{基础情况}

当$n = 2$时，显然有
\[ \left(\sqrt{a_1} - \sqrt{a_2}\right)^2 \ge 0 \]
展开后立得
\[ a_1 + a_2 \ge 2\sqrt{a_1a_2} \]
当且仅当$a_1 = a_2$时等号成立。

\subsubsection{递归情况}

若对于$1 < k \le n$有
\[ \sum_{i = 1}^k a_i \ge k\sqrt[k]{\prod_{i = 1}^k a_i} \]
成立，则对于$n + 1$，令
\[ S = \sum_{i = 1}^n a_i, P = \prod_{i = 1}^n a_i, S' = \sum_{i = 1}^{n + 1} a_i, P' = \prod_{i = 1}^{n + 1} a_i \]
有
\begin{align*}
  S' &= S + \left(a_{n + 1} + (n - 1)\sqrt[n + 1]{P'}\right) - (n - 1)\sqrt[n + 1]{P'} \\
  &\ge n\sqrt[n]{P} + n\sqrt[n]{a_{n + 1}\left(\sqrt[n + 1]{P'}\right)^{n - 1}} - (n - 1)\sqrt[n + 1]{P'} \\
  &\ge 2n\sqrt{\sqrt[n]{P}\sqrt[n]{a_{n + 1}\left(\sqrt[n + 1]{P'}\right)^{n - 1}}} - (n - 1)\sqrt[n + 1]{P'} \\
  &= 2n\sqrt[2n]{a_{n + 1}P\left(\sqrt[n + 1]{P'}\right)^{n - 1}} - (n - 1)\sqrt[n + 1]{P'} \\
  &= 2n\sqrt[2n]{\left(\sqrt[n + 1]{P'}\right)^{n + 1}\left(\sqrt[n + 1]{P'}\right)^{n - 1}} - (n - 1)\sqrt[n + 1]{P'} \\
  &= 2n\sqrt[2n]{\left(\sqrt[n + 1]{P'}\right)^{2n}} - (n - 1)\sqrt[n + 1]{P'} \\
  &= (2n - (n - 1))\sqrt[n + 1]{P'} = (n + 1)\sqrt[n + 1]{P'} \\
\end{align*}
当且仅当
\begin{align*}
  a_1 = a_2 &= \dots = a_n \\
  a_{n + 1} &= \sqrt[n + 1]{P'} \\
  \sqrt[n]{P} &= \sqrt[n]{a_{n + 1}\left(\sqrt[n + 1]{P'}\right)^{n - 1}}
\end{align*}
即$a_1 = a_2 = \dots = a_{n + 1}$时等号成立。原不等式得证。
